DFS——Red and Black

最新推荐文章于 2023-08-07 16:49:46 发布

Chiac_az

最新推荐文章于 2023-08-07 16:49:46 发布

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分类专栏：深度优先搜索文章标签：算法 dfs c语言

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本文介绍了一个基于深度优先搜索（DFS）算法的迷宫寻路问题解决方案。在一个由红色和黑色方块组成的矩形房间中，从一个指定的黑色方块出发，寻找所有可达的黑色方块路径。通过递归地探索四个相邻的黑色方块，避免红色障碍，最终统计并返回可到达的所有黑色方块数量。文章提供了完整的C语言实现代码，展示了如何有效利用DFS算法解决此类问题。

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题目

There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can’t move on red tiles, he can move only on black tiles.

Write a program to count the number of black tiles which he can reach by repeating the moves described above.
Input
The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.

There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.

‘.’ - a black tile
‘#’ - a red tile
‘@’ - a man on a black tile(appears exactly once in a data set)
Output
For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).

Sample Input
6 9
…#.
…#
…
…
…
…
…
#@…#
.#…#.
11 9
.#…
.#.#######.
.#.#…#.
.#.#.###.#.
.#.#…@#.#.
.#.#####.#.
.#…#.
.#########.
…
11 6
…#…#…#…
…#…#…#…
…#…#…###
…#…#…#@.
…#…#…#…
…#…#…#…
7 7
…#.#…
…#.#…
###.###
…@…
###.###
…#.#…
…#.#…
0 0

Sample Output
45
59
6
13
与lake counting题结构相似，不过只需从一个地方开始。下面代码可以将输入与找@合并，scanf放在循环里。

#include<stdio.h>
#include<string.h>
int w,h,tx,ty,cnt=1,max=0;		//w为行数，h为列数
char a[27][27],flag[27][27];	//flag记录该位置有没被访问过
int next[4][2]={				//四个方向
 1,0,
 0,1,
-1,0,
 0,-1
};
void dfs(int x,int y){			//递归函数
	for(int v=0;v<4;v++){		//在每个地方都有四个方向可走
		int dx=x+next[v][0];
		int dy=y+next[v][1];
		if(a[dx][dy]=='.'&&dx>0&&dx<=w&&dy>0&&dy<=h&&!flag[dx][dy]){
			flag[dx][dy]=1;
			dfs(dx,dy);
		}
	}
}
int main(){
	scanf("%d %d",&h,&w);
	while(h&&w){			//都不为0
		getchar();
		for(int i=1;i<=w;i++){//输入数据
			for(int j=1;j<=h;j++){
				scanf("%c",&a[i][j]);
			}
			getchar();
		}
		memset(flag,0,sizeof(flag));
		for(int i=1;i<=w;i++){
			for(int j=1;j<=h;j++){
				if(a[i][j]=='@'){	//从@开始
					tx=i;
					ty=j;
				}
			}
		}
		dfs(tx,ty);
		for(int i=1;i<=w;i++){		//数flag表里有几个1
			for(int j=1;j<=h;j++){
				if(flag[i][j]==1){
					cnt++;
				}
			}
		}
/*		for(int i=1;i<=w;i++){		//输出flag表
			for(int j=1;j<=h;j++){
				printf("%2.d",flag[i][j]);
			}
			printf("\n");
		}*/
		printf("%d\n",cnt);
		cnt=1;
		scanf("%d %d",&h,&w);
	}
	return 0;
}